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Sagot :
I can't imagine that this is going to do you much good, but
I'm sure going to enjoy solving it.
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Skip this whole first section.
It was an attempt to master a bunch of trees, while
the forest was right there in front of me all the time.
Drop down below the double line.
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Kepler's 3rd law says:
(square of the orbital period) / (cube of the orbital radius) = constant
T₀² = K R₀³
I put the zero subscripts in there, because you doubled 'R'
and I need to know how that affected 'T'.
new-T² = K(2 R₀)³
new-T² = 8 K (R₀)³ = 8 old-T₀²
new-T = √8 old-T <=== that's what I was after
I just teased out the Moon's new orbital period if it's distance were doubled.
Instead of 1 month, it's now √8 months.
To put a somewhat sharper point on it, the moon's period of revolution
changes from 27.322 days to 27.322√8 = 77.278 days (rounded) .
Using 385,000 km for the moon's current average distance, the current orbital speed is
(2π x 385,000 km) / (27.322 days) = 1,024.7 m/s
(One online source says 1.023 km, so we're not doing too badly so far.)
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I'm such a dummy. I don't need to go through all of that.
If the moon were twice as far from Earth as it really is, then it would
average 770,000 km instead of the present 385,000 km.
That's 120.86 times the Earth's radius of 6,371 km.
So the acceleration of gravity out there would be
(1 / 120.86)² of the (9.807 m/s²) that it is here on the surface.
new-G = 0.000671 m/s²
Distance a dropped object falls = 1/2 g t²
In the first second, that's 1/2 g (1)² = 1/2 g
For an orbiting object, every second is the "first"second, because ...
as we often explain orbital motion qualitatively ... the Earth "falls away"
just as fast as the curved orbit falls.
Distance an object falls in the 1st second =
1/2 G = 0.000336 m/s = 0.336 millimeter per second
I estimate the probability of a mistake somewhere during this process
at approx 99.99% . But I don't have anything better right now, and I've
wasted too much time on it already, so I'll stick with it.
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